MOTION IN ONE DIMENSION

AVERAGE / INSTANTANEOUS SPEED

POSITION AND DISPLACEMENT

AVERAGE / INSTANTANEOUS VELOCITY

AVERAGE / INSTANTANEOUS ACCELERATION

MOTION WITH CONSTANT ACCELERATION

FREE FALL

P

P

P

P

P

P

KINEMATICS is the study of motion in time and space withoutconsidering the cause of the motion.

PPOSITION is the location of an object with reference to acoordinate system. In one dimension the coordinate x isusually used.

PWhen an object is stationary (at rest) its position does notchange. When an object is moving its position does change.

PDuring a specific time interval in which motion occurs theobject moves from an initial position xi at the beginning of thetime interval to a final position xf at the end of the timeinterval.

PThe beginning and end of the time interval are designated t i(initial) and tf (final) respectively.

DISPLACEMENT is the change in position of an object. Displacement is designated x.

A VECTOR is a physical quantity that has both size (magnitude)and direction. Physical quantities that do not have direction areSCALARS.

Position and displacement are both vectors.

For vectors in one dimension, there are only two directions;toward + or toward -. If the vector points toward +, thevector is positive, like +6m. If the vector points toward -, thevector is negative, like -6m. Thus the + and - signs are used todesignate direction for one dimensional vectors.

f ix x x

If the wheel moves from x = -1 to x = 3 and then to x = -2, what overall displacement has the wheel experienced?

-4 -3 -2 -1 0 1 2 3 4

Start ReverseFinishEXAMPLE:

If the wheel moves from x = -1 to x = 3 and then to x = -2, what overall displacement has the wheel experienced?

( 1 1( 2) )f ix x x

-4 -3 -2 -1 0 1 2 3 4

Start ReverseFinishANSWER

Average velocity is designated and is defined as displacement, x,divided by the time over which the displacement occurs, t.

v

How is average velocity different from average speed?

Average speed is the total distance traveled divided by the total traveltime. Since distance is always positive and has no direction, averagespeed also is always positive and has no direction. Average velocitydoes depend on direction and may be positive or negative dependingon direction. Usually an object’s average speed isn’t numericallyequal to its average velocity.

Do not use the word speed when you mean velocity.

f i

f i

x xxv

t t t

t

x

A

B

xi

xf

tfti

x

t

Slope of secant line between points A and Bequals average velocity.

Red curve shows a plot of position vs time foran object in motion

f i f ixv x x t tt

The INSTANTANEOUS VELOCITY, designated v, of amoving object is its velocity determined at an instant in time.

The instantaneous speed of an object is the magnitude of itsinstantaneous velocity. No other relationship betweenspeed and velocity exists.

0lim

t

xv

t

t

x Red curve shows a plot of position vs time foran object in motion

Slope of tangent line equals theinstantaneous velocity of theobject at that point on the graph.

x

t

xv

t

If the wheel moves from x = -1m at t = 0 to x = 4m at t = 3s, and then to x = -3m at t = 8s, what are the overall average speed and average velocity of the wheel?

-4 -3 -2 -1 0 1 2 3 4

Start ReverseFinish

EXAMPLE:

If the wheel moves from x = -1m at t = 0 to x = 4m at t = 3s, and then to x = -3m at t = 8s, what are the overall average speed and average velocity of the wheel?

total distance 12Avg speed

elapsed time 8

displacement ( 3 ) ( 1 )Avg vel

elapsed tim

1.5

.e 8

25

ms

ms

m

s

m m

s

-4 -3 -2 -1 0 1 2 3 4

Start ReverseFinish

ANSWER

The AVERAGE ACCELERATION of an object in motion, , is its change in velocity divided by the time it takes forthe change to occur.a

Effects of Acceleration:

P When an object’s velocity and acceleration are in thesame direction, the object speeds up.

P When velocity and acceleration are in oppositedirections, the object slows down.

P When an object changes its direction of motion, only itsvelocity vector changes direction (or sign). At theposition of the change in direction, velocity isinstantaneously zero but acceleration is non-zero.

f i

f i

v vva

t t t

Red curve is graph of instantaneous velocity vs timefor an object in motion

v

t

vf

vi

ti tf

v

t

Slope of secant line is the average accelerationoccurring between points A and B.

A

B

a v t

INSTANTANEOUS ACCELERATION is the accelerationan object has at a certain time, for an instant. It is definedas:

Displacement is change in position.

Velocity is the rate of change in position.

Acceleration is the rate of change in velocity.

0lim

t

va

t

Red curve is graph of instantaneous velocity vs timefor an object in motion

v

t

v

t

Slope of tangent line is the instantaneousacceleration occurring at that point.

a v t

MOTION IN ONE DIMENSION W ITH CONSTANT ACCELERATION.

W hen acceleration is constant: a a

It is convenient to let v i = v0 and v f = v so that v = v - vo

and ti = 0 and tf = t so that t = t.

00

v vva v v at

t t

210

0

00

0 00

20

0

0 2

2

2

2

2

2

ov v x xxv

t tvt v t

x x

v at t v tx x

v t atx

x

x

x v t at

Also let xi = x0 and xf = x so that x = x - x0.

00

0102

2 20

2 20 0

2

2

v vv v at t

av v

x vt v va

v v a

v

a

x

v

x

x

Kinematic Equations In One Dimension With Constant Accelerations

210 0 2

0

2 20 0

0 0

2

2

x x v t at

v v at

v v a x x

x x v vv

t

EXAMPLE: While being thrown a baseball moves from rest to a speed of about 90 mph in distance about that of 1.5m. Calculate the acceleration of such a baseball.

ANSWER: While being thrown a baseball moves from rest to a speed of about 90 mph in distance about that of 1.5m. Calculate the acceleration of such a baseball.

2

16093600

2 2

2 2 2 2

90 90 40.225

2 solve for a

(40.225 ) 0

2 2 1 539.35

.5

miles m hour mhour mile s

m

s

f

mi

s

i

f s

mph

v v a x

v va

xa

m

Objects in free-fall exhibit motion with constant acceleration.

Conditions for free-fall:

PAir resistance is small enough to ignore.

P The acceleration due to gravity may be treated asuniform throughout the motion of the object.

P Effects of Earth’s rotation and curvature may alsobe ignored.

An object in free-fall moves vertically so we switch to avertical axis and denote position with the variable y.

The magnitude of the free-fall acceleration is the constantg = 9.80 m/s2

Before the kinematic equations can be used in a free-fallproblem the coordinate axis must be oriented.

If + is upward, then a = -g. All vectors pointing upwardare positive and all pointing downward are negative.

If + is downward, then a = +g. All vectors pointing upwardare negative and all pointing downward are positive.

In all cases the acceleration due to free-fall is 9.80m/s2 in thedownward direction.

The kinematic equations for free-fall:

210 0 2

0

2 20 0

0 0

2

2

y y v t gt

v v gt

v v g y y

y y v vv

t

210 0 2

0

2 20 0

0 0

2

2

y y v t gt

v v gt

v v g y y

y y v vv

t

If the positive y-axis pointsdownward:

If the positive y-axis pointsupward:

EXAMPLE: A ball is thrown vertically upward near the edge of the roof of a tall building 60m above the ground. The initial upward speed of the ball is 12m/s. On the way down, the ball will just miss the edge of the building. (a) What maximum height does the ball reach? (b) How long is the ball in the air? (c) What is the speed of the ball when it strikes the ground?

ANSWER: A ball is thrown vertically upward near the edge of the roof of a tall building 60m above the ground. The initial upward speed of the ball is 12m/s. On the way down, the ball will just miss the edge of the building. (a) What maximum height does the ball reach? (b) How long is the ball in the air? (c) What is the speed of the ball when it strikes the ground?

2

2

2 2i f max f

2 2 2

max

2 2

2 ( ) with y 60 , y y , v 0

0 (12 )60

2 2 9.8

2 ( ) (12 ) 2 9.8 (0 60 )

(answer to

67.347

36.332 part c)

solve for

t

f i f i

mf i s

i ms

m mf i f i s s

f i

f i

mf s

v v g y y m

v vy y m

g

v v g y y m

v v gt

v v

m

t

v

2

36.332 14.93

2

9.2

8

m ms s

ms

gs